I'm not sure how to show this:
Show the field $\mathbb{F}_4[X] /(X^2 +X + \alpha)$ is isomorphic to the field $\mathbb{F}_{16}=\mathbb{F}_2[x]/(x^4 + x^3 + 1)$ by explicitly constructing an isomorphism between the two fields. Let $\alpha$ be the element (0 1) in $\mathbb{F}_4$, i.e., $\alpha$ = $x$ in polynomial form.
Any help or hint would be helpful.
Writing
is more confusing than anything. Just write $\alpha$ and explicitly state that $\mathbb{F}_4$ has teh elements $0,1,\alpha, \alpha+1$ or the likes, it is clearer.
Now onto the question, the question is about showing an explicit homomorphism, let $\beta$ be the root to $X^2+X+\alpha=0$, as such we have all elements in $\Bbb F_4/(X^2+X+\alpha)$ in the form of $n\beta+m$ with $m,n\in\Bbb F_4$. Which we can make into $$n_1+n_2\alpha+n_3\beta+n_4\alpha\beta$$ with $n_i\in\Bbb F_2$. From this do we have a basis and now we just to mape it properly. For $\Bbb F_{16}$ we have $\gamma^i$ being the basis, with $\gamma^4=\gamma+1$. Then we just need to find which element has that property and map it to our $\gamma$. A quick check gives us that $\beta^4=\beta+1$ and as such we have that $\beta\mapsto\gamma$. Next we remember that $\beta^2=\beta+\alpha$ which gives $\beta^2+\beta=\alpha$ and as such $\alpha\mapsto\gamma^2+\gamma$ and finally $\alpha\beta\mapsto\gamma^3+\gamma^2$.
You need to show this is a homomorphism. We can do this in iterations of all permutations which I do for simplicity, Addition is trivial and I'll do one of the multiplication pairs. $$\varphi(\alpha\cdot\alpha\beta)=\varphi(\alpha\beta+\beta)=\gamma^3+\gamma^2+\gamma=$$ $$\gamma^3+\gamma(\gamma+1)=\gamma^3\gamma^4+\gamma^5+\gamma^4=(\gamma^2+\gamma)(\gamma^3+\gamma^2)=\varphi(\alpha)\cdot\varphi(\alpha\beta)$$
Alternatively can you show that the powers of $\beta$ gives the others.
After that it is trivial showing it is surjective and injective.